(Resisor-Capacior Circuis AP Physics C
Circui Iniial Condiions An circui is one where you have a capacior and resisor in he same circui. Suppose we have he following circui: Iniially, he capacior is UNCHARGED ( 0 and he curren hrough he resisor is zero. A swich (in red hen closes he circui by moving upwards. The uesion is: Wha happens o he curren and volage across he resisor and capacior as he capacior begins o charge as a funcion of ime? Which pah do you hink i akes? V C Time(s
Volage Across he Resisor - Iniially V Resisor ε If we assume he baery has NO inernal resisance, he volage across he resisor will be he EMF. (sec Afer a very long ime, V cap ε, as a resul he poenial difference beween hese wo poins will be ZERO. Therefore, here will be NO volage drop across he resisor afer he capacior charges. Noe: This is while he capacior is CHARGING.
Curren Across he Resisor - Iniially I max ε/r (sec Since he volage drop across he resisor decreases as he capacior charges, he curren across he resisor will reach ZERO afer a very long ime. Noe: This is while he capacior is CHARGING.
Volage Across he Capacior - Iniially V cap ε (sec As he capacior charges i evenually reaches he same volage as he baery or he EMF in his case afer a very long ime. This increase DOES NOT happen linearly. Noe: This is while he capacior is CHARGING.
Curren Across he Capacior - Iniially I max ε/r (sec Since he capacior is in SERIES wih he resisor he curren will decrease as he poenial difference beween i and he baery approaches zero. I is he poenial difference which drives he value for he curren. Noe: This is while he capacior is CHARGING.
Time Domain Behavior The graphs we have jus seen show us ha his process depends on he ime. Le s look hen a he UNITS of boh he resisance and capaciance. Uni for Resisance Ω Vols/Amps Uni for Capaciance Farad Coulombs/Vols Vols Coulombs R xc x Amps Vols Coulomb 1 Amp Sec Coulombs R xc Coulombs Seconds Coulombs Amps SECONDS!
The Time Consan I is clear, ha for a GIVEN value of "C, for any value of R i effecs he ime rae a which he capacior charges or discharges. Thus he PRODUCT of R and C produce wha is called he CIUIT Capaciive TIME CONSTANT. We use he Greek leer, Tau, for his ime consan. The uesion is: Wha exacly is he ime consan?
The Time Consan The ime consan is he ime ha i akes for he capacior o reach 63% of he EMF value during charging.
Charging Behavior Is here a funcion ha will allow us o calculae he volage a any given ime? ε Le s begin by using KVL V cap ε (sec We now have a firs order differenial euaion.
Charging funcion ε How do we solve his when we have 2 changing variables? To ge rid of he differenial we mus inegrae. To make i easier we mus ge our wo changing variables on differen sides of he euaion and inegrae each side respecively. Re-arranging algebraically. Geing he common denominaor Separaing he numeraor from he denominaor, Cross muliplying. Since boh changing variables are on opposie side we can now inegrae.
Charging funcion 0 d Cε Cε ln( Cε C ε e Cε Cε Cε e 1 0 d ( Cε Cε e Cε (1 e ε However if we divide our funcion by a CONSTANT, in his case C, we ge our volage funcion. As i urns ou we have derived a funcion ha defines he CHARGE as a funcion of ime. ( C ε (1 e C C V ( ε (1 e
Le s es our funcion V V V ( ε (1 e (1 ε (1 e (1 ε (1 e 1 0.63ε ε 0.95ε 0.98ε 0.86ε 0.63ε Transien Sae Seady Sae V (2 0.86ε V V (3 (4 0.95ε 0.98ε 1 2 31 4 Applying each ime consan produces he charging curve we see. For pracical purposes he capacior is considered fully charged afer 4-5 ime consans( seady sae. Before ha ime, i is in a ransien sae.
Charging Funcions ( Cε (1 e V ( ε (1 e V ( R ε (1 e R I ( Io (1 e o Likewise, he volage funcion can be divided by anoher consan, in his case, R, o derive he curren charging funcion. Now we have 3 funcions ha allows us o calculae he Charge, Volage, or Curren a any given ime while he capacior is charging.
Capacior Discharge Resisor s Volage Suppose now he swich moves downwards owards he oher erminal. This prevens he original EMF source o be a par of he circui. ε V Resisor A 0, he resisor ges maximum volage bu as he capacior canno keep is charge, he volage drop decreases. (sec
Capacior Discharge Resisor s Curren Similar o is charging graph, he curren hrough he resisor mus decrease as he volage drop decreases due o he loss of charge on he capacior. Iε/R I Resisor (sec
Capacior Discharge Capacior's Volage The discharging graph for he capacior is he same as ha of he resisor. There WILL be a ime delay due o he TIME CONSTANT of he circui. In his case, he ime consan is reached when he volage of he capacior is 37% of he EMF.
Capacior Discharge Capacior s Curren Similar o is charging graph, he curren hrough he capacior mus decrease as he volage drop decreases due o he loss of charge on he capacior. Iε/R I cap (sec
Discharging Funcions 0 IR V cap d R d C d R d C o 1 d 0 1 0 d d 0 d Once again we sar wih KVL, however, he reason we sar wih ZERO is because he SOUE is now gone from he circui.
Discharging Funcions o 1 d 1 0 d ln( o o e ( Dividing by "C" hen "R" o e V ( I( ε e I o o e We now can calculae he charge, curren, or volage for any ime during he capaciors discharge.